Question: $\int (6x^2-5x+3)\,dx=$ $+C$
Explanation: We can use the sum rule and the constant multiple rule for indefinite integrals: $\begin{aligned} &\int [f(x)+g(x)]dx=\int f(x)\,dx+\int g(x)\,dx \\\\\\ &\int k\cdot f(x)= k\cdot\int f(x)\,dx \end{aligned}$ Using the sum and the constant multiple rules, we can rewrite our integral as follows: $\int (6x^2-5x+3)\,dx=6\int x^2\,dx-5\int x\,dx+3\int 1\,dx$ Now we can find each indefinite integral using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ Note: we can only use the reverse power rule because $n \neq -1$. $\begin{aligned} &\phantom{=}\int (6x^2-5x+3)\,dx \\\\ &=6\int x^2\,dx-5\int x\,dx+3\int 1\,dx \\\\ &=6\dfrac{x^3}{3}-5\dfrac{x^2}{2}+3\dfrac{x^1}{1}+C \\\\ &=2x^3-\dfrac52x^2+3x+C \end{aligned}$ In conclusion, $\int (6x^2-5x+3)\,dx=2x^3-\dfrac52x^2+3x+C$